Question: The angle $\theta_1$ is located in Quadrant $\text{IV}$, and $\cos(\theta_1)=\dfrac{3}{5}$. What is the value of $\sin(\theta_1)$ ? Express your answer exactly. $\sin(\theta_1)=$
Solution: The Strategy We can use the Pythagorean identity, $\cos^2(\theta)+\sin^2(\theta)=1$, to solve for $\cos(\theta_1)$ from $\sin(\theta_1)$ or vice versa. [How did we get the Pythagorean identity?] In this case, we can find $\sin(\theta_1)$ by doing the following. Find $\sin^2(\theta_1)$ using $\cos(\theta_1)$ and the Pythagorean identity. [What does this notation mean?] Determine $\sin(\theta_1)$ by considering the quadrant of $\theta_1$. Finding $\sin^2(\theta_1)$ Let's plug in $\cos(\theta_1)=\dfrac{3}{5}$ into the equation $\cos^2(\theta)+\sin^2(\theta)=1$ to solve for $\sin^2(\theta_1)$. $\begin{aligned}\cos^2(\theta_1)+\sin^2(\theta_1)&=1 \\\\\sin^2(\theta_1)&={1-\cos^2(\theta_1)} \\&=1-\left(\dfrac{3}{5}\right)^2 \\&=\dfrac{16}{25}\end{aligned}$ Finding $\sin(\theta_1)$ Since $\theta_1$ is in Quadrant $\text{IV}$, $\sin(\theta_1)$ is negative. $\begin{aligned}\sin(\theta_1)&=-\sqrt{\sin^2(\theta_1)} \\&=-\sqrt{\dfrac{16}{25}} \\&=-\dfrac{4}{5}\end{aligned}$ Summary $\sin(\theta_1)=-\dfrac{4}{5}$